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Pedagogically, the techniques needed to handle general coordinate systems (aka non-inertial frames) are taught in GR courses, but as far as the physics go, if you have flat space time (i.e. no gravity), you can use the techniques of SR just fine. You could conceivable even do the analysis without tensors, though if you want to compare your results to textbook results, I'm not aware of any textbooks that don't use tensors (not that I've read them all).

A brief outline of one way to go about doing this:

1) Solve the relativistic rocket equation for a constant proper acceleration rocket. Check your solution against the standard ones on wikipedia and/or the relativistic rocket FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken].

I will use the following notation: let your reference inertial frame be (t,x,y,z). Let your rocket coordinates be (T,X,Y,Z). Let the momentarily co-moving inertial frame coordinates at some proper time T be represented by (t', x', y', z'). Let the acceleration of the rocket be called g.

So solving the relativistic rocket equation you want to find z(T) and t(T), the inertial coordinate frame position and time as a function of rocket time T, which is just the proper time of the accelerating rocket.

2) Consider the momentarily co-moving inertial frame at proper time T on the rocket. Because of linearity, there will be a linear relationship between ##\hat{z'}##,##\hat{t'}##, the components of the basis vectors in the momentarily comoving inertial frame at time T, and ##\hat{z}##, ##\hat{t}## the basis vectors in the inertial frame. Basically all we really need to do is figure out the components of ##\hat{z'}## in the inertial frame, this vector will have some time components due to the relativity of simultaneity, and some space-component due to relativistic length contraction

3) Find an expression that converts rocket coordinates (T,X,Y,Z) to inertial coordinates (t,x,y,z). We will basically have

t = t(T) + Z * (t-component of ##\hat{z'}##)

x = X

y = Y

z = z(T) + Z * (z-component of ##\hat{z'}##)

here t(T) and z(T) are the expressions we derived in part 1, and while the relationship between changes in the Z coordinate at time T and the changes in the t and z coordinates is linear, we need to find out what the coefficients are.

At this point we've codified precisely what we mean by "rocket coordinates" by having an expression that gives the inertial coordinates (t,x,y,z) as functions of the rocket coordinates (T,X,Y,Z).

4) Optional. Using the above results from 3, find the metric in the accelerating coordinates by taking dx^2 + dy^2 + dz^2 - c^2 dt^2 in terms of dT,dX,dY, and dZ. This is just algebra (albeit lengthly without computer assistance). Compare it to the textbook results, which say that you should get dX^2 + dY^2 + dZ^2 - c^2 (1+gZ)^2 dT^2

5) Optional, but recommened. Consider whether the mapping derived in part 3 is a 1:1 mapping (hint: we need to insist that 1+gZ is positive if we want to have a 1:1 mapping).

A brief outline of one way to go about doing this:

1) Solve the relativistic rocket equation for a constant proper acceleration rocket. Check your solution against the standard ones on wikipedia and/or the relativistic rocket FAQ at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken].

I will use the following notation: let your reference inertial frame be (t,x,y,z). Let your rocket coordinates be (T,X,Y,Z). Let the momentarily co-moving inertial frame coordinates at some proper time T be represented by (t', x', y', z'). Let the acceleration of the rocket be called g.

So solving the relativistic rocket equation you want to find z(T) and t(T), the inertial coordinate frame position and time as a function of rocket time T, which is just the proper time of the accelerating rocket.

2) Consider the momentarily co-moving inertial frame at proper time T on the rocket. Because of linearity, there will be a linear relationship between ##\hat{z'}##,##\hat{t'}##, the components of the basis vectors in the momentarily comoving inertial frame at time T, and ##\hat{z}##, ##\hat{t}## the basis vectors in the inertial frame. Basically all we really need to do is figure out the components of ##\hat{z'}## in the inertial frame, this vector will have some time components due to the relativity of simultaneity, and some space-component due to relativistic length contraction

3) Find an expression that converts rocket coordinates (T,X,Y,Z) to inertial coordinates (t,x,y,z). We will basically have

t = t(T) + Z * (t-component of ##\hat{z'}##)

x = X

y = Y

z = z(T) + Z * (z-component of ##\hat{z'}##)

here t(T) and z(T) are the expressions we derived in part 1, and while the relationship between changes in the Z coordinate at time T and the changes in the t and z coordinates is linear, we need to find out what the coefficients are.

At this point we've codified precisely what we mean by "rocket coordinates" by having an expression that gives the inertial coordinates (t,x,y,z) as functions of the rocket coordinates (T,X,Y,Z).

4) Optional. Using the above results from 3, find the metric in the accelerating coordinates by taking dx^2 + dy^2 + dz^2 - c^2 dt^2 in terms of dT,dX,dY, and dZ. This is just algebra (albeit lengthly without computer assistance). Compare it to the textbook results, which say that you should get dX^2 + dY^2 + dZ^2 - c^2 (1+gZ)^2 dT^2

5) Optional, but recommened. Consider whether the mapping derived in part 3 is a 1:1 mapping (hint: we need to insist that 1+gZ is positive if we want to have a 1:1 mapping).

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